 Nann

2021-02-24

A study of young children was designed to increase their intake of whole-grain, rather than regular-grain, snacks. At the end of the study, the 79 children who participated in the study were presented with a choice between a regular-grain snack and a whole-grain alternative. The whole-grain alternative was chosen by 55 children. You want to examine the possibility that the children are equally likely to choose each type of snack.
${H}_{0}:p=0.5$
${H}_{a}:p\ne 0.5$
Perform the significance test. (Use $\alpha =0.01$. Round your test statistic to two decimal places and your P-value to four decimal places.) rogreenhoxa8

Step 1
Given Information
No of children participated $\left(n\right)=79$
No of children chosen $\left(x\right)=55$
$\stackrel{^}{p}=\frac{x}{n}=\frac{55}{79}=0.696$
Null and alternative Hypothesis
Null Hypothesis: $p=0.5$
Alternative Hypothesis: $p\ne 0.5$
Test statistic
$z=\frac{\stackrel{^}{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}=\frac{0.696-0.5}{\sqrt{\frac{0.5×\left(1-0.5\right)}{79}}}=3.48}$
Step 2
p-value
$p-value=2xP\left(Z>3.48\right)=2x\left(1–P\left(Z<3.48\right)\right)=2x\left(1–0.999748\right)=0.000501$

$p-value<0.01$,
So, reject null hypothesis and conclude that children are not equally likely to choose each type of snack.

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